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A $15\mu F$ capacitor is connected to 220 V, 50 Hz source. Find the peak current.

Last updated date: 03rd Mar 2024
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Hint The voltage is non constant, and can be considered as a sine function. When voltage value is stated, the root mean square value is used. The peak value is equal to the root means squared value multiplied by the square root of two.
In this solution we will be using the following formula;
$\Rightarrow i = C\dfrac{{dv}}{{dt}}$
Where $i$ is the instantaneous current at a particular time $t$ on a capacitor, $C$ is the capacitance of the capacitor and $v$ is the instantaneous voltage across the capacitor at the same corresponding time $t$.

Complete step by step answer
In the question, we have that a capacitor is connected to 220 V, 50 Hz source. Hence, mathematically, the voltage can be given as,
$\Rightarrow v = {V_p}\sin ft = {v_{rms}}\sqrt 2 \sin 2\pi ft$
$\Rightarrow v = 220\sqrt 2 \sin 2\pi \left( {50} \right)t$
Hence, by differentiating, we have that
$\Rightarrow \dfrac{{dv}}{{dt}} = 11000 \times 2\pi \sqrt 2 \cos 2\pi \cdot 50t$
The instantaneous current in the presence of a capacitor is given as
$\Rightarrow i = C\dfrac{{dv}}{{dt}}$, where $C$ is the capacitance of the capacitor.
Hence substituting the voltage expression from above and the capacitance value from the question, we have that
$\Rightarrow i = 15 \times {10^{ - 6}} \times 11000 \times 2\pi \sqrt 2 \cos 2\pi \cdot 50t$
Hence, by computation, we have
$\Rightarrow i = 1.47\cos 2\pi \cdot 50t$
Hence, the peak value of the current is ${\text{1}}{\text{.47A}}$.

For clarity, we shall derive the expression of the instantaneous current, $i = C\dfrac{{dv}}{{dt}}$.
In general, the charge in a capacitor is defined simply as
$\Rightarrow Q = Cv$. The current flowing through the capacitor is hence given as
$\Rightarrow i = \dfrac{{dQ}}{{dt}}$, thus from the expression of the charge above, we have that
$\Rightarrow i = \dfrac{{dQ}}{{dt}} = \dfrac{{d\left( {Cv} \right)}}{{dt}} = C\dfrac{{dv}}{{dt}}$ (since the capacitance is a constant)
Hence, finally
$\Rightarrow i = C\dfrac{{dv}}{{dt}}$
Also, capacitors only allow AC current to flow through them. As we see, if the voltage is a constant, then
$\Rightarrow \dfrac{{dv}}{{dt}} = 0$
Hence, we have that
$\Rightarrow i = C\dfrac{{dv}}{{dt}} = 0$.
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